Why does the Lagrange multiplier $lambda$ change when the equality constraint is scaled?












2












$begingroup$


Consider the problem



$$begin{array}{ll} text{maximize} & x^2+y^2 \ text{subject to} & dfrac{x^2}{25} + dfrac{y^2}{9} = 1end{array}$$



Solving this using the Lagrange multiplier method, I get



$$x = pm5, qquad y = 0, qquad lambda = 25$$



However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?






optimization lagrange-multiplier constraints qcqp







share|cite|improve this question











$endgroup$












  • $begingroup$
    That shouldn't be the case - maybe check for an arithmetic error?
    $endgroup$
    – Vasting
    22 hours ago










  • $begingroup$
    Double checked, doesn't seem to be an error.
    $endgroup$
    – PGupta
    21 hours ago






  • 1




    $begingroup$
    Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
    $endgroup$
    – littleO
    21 hours ago










  • $begingroup$
    Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
    $endgroup$
    – PGupta
    21 hours ago
















2












$begingroup$


Consider the problem



$$begin{array}{ll} text{maximize} & x^2+y^2 \ text{subject to} & dfrac{x^2}{25} + dfrac{y^2}{9} = 1end{array}$$



Solving this using the Lagrange multiplier method, I get



$$x = pm5, qquad y = 0, qquad lambda = 25$$



However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?






optimization lagrange-multiplier constraints qcqp







share|cite|improve this question











$endgroup$












  • $begingroup$
    That shouldn't be the case - maybe check for an arithmetic error?
    $endgroup$
    – Vasting
    22 hours ago










  • $begingroup$
    Double checked, doesn't seem to be an error.
    $endgroup$
    – PGupta
    21 hours ago






  • 1




    $begingroup$
    Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
    $endgroup$
    – littleO
    21 hours ago










  • $begingroup$
    Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
    $endgroup$
    – PGupta
    21 hours ago














2












2








2





$begingroup$


Consider the problem



$$begin{array}{ll} text{maximize} & x^2+y^2 \ text{subject to} & dfrac{x^2}{25} + dfrac{y^2}{9} = 1end{array}$$



Solving this using the Lagrange multiplier method, I get



$$x = pm5, qquad y = 0, qquad lambda = 25$$



However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?






optimization lagrange-multiplier constraints qcqp







share|cite|improve this question











$endgroup$




Consider the problem



$$begin{array}{ll} text{maximize} & x^2+y^2 \ text{subject to} & dfrac{x^2}{25} + dfrac{y^2}{9} = 1end{array}$$



Solving this using the Lagrange multiplier method, I get



$$x = pm5, qquad y = 0, qquad lambda = 25$$



However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?






optimization lagrange-multiplier constraints qcqp




optimization lagrange-multiplier constraints qcqp






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









Rodrigo de Azevedo

13.1k41962




13.1k41962










asked 22 hours ago









PGuptaPGupta

1746




1746












  • $begingroup$
    That shouldn't be the case - maybe check for an arithmetic error?
    $endgroup$
    – Vasting
    22 hours ago










  • $begingroup$
    Double checked, doesn't seem to be an error.
    $endgroup$
    – PGupta
    21 hours ago






  • 1




    $begingroup$
    Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
    $endgroup$
    – littleO
    21 hours ago










  • $begingroup$
    Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
    $endgroup$
    – PGupta
    21 hours ago


















  • $begingroup$
    That shouldn't be the case - maybe check for an arithmetic error?
    $endgroup$
    – Vasting
    22 hours ago










  • $begingroup$
    Double checked, doesn't seem to be an error.
    $endgroup$
    – PGupta
    21 hours ago






  • 1




    $begingroup$
    Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
    $endgroup$
    – littleO
    21 hours ago










  • $begingroup$
    Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
    $endgroup$
    – PGupta
    21 hours ago
















$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
22 hours ago




$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
22 hours ago












$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
21 hours ago




$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
21 hours ago




1




1




$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
21 hours ago




$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
21 hours ago












$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
21 hours ago




$begingroup$
Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
$endgroup$
– PGupta
21 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$



Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
$$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$



The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$



If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$



and we'll get the same answer of course.



It happens because $-frac{1}{9}cdot225=-25$ and
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
    $endgroup$
    – PGupta
    21 hours ago










  • $begingroup$
    @PGupta I added something. See now.
    $endgroup$
    – Michael Rozenberg
    21 hours ago










  • $begingroup$
    This makes sense, thank you.
    $endgroup$
    – PGupta
    21 hours ago



















4












$begingroup$

If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag{1} nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.



Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
    $endgroup$
    – PGupta
    21 hours ago






  • 1




    $begingroup$
    @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
    $endgroup$
    – littleO
    20 hours ago










  • $begingroup$
    Great, got it. Thanks a lot!
    $endgroup$
    – PGupta
    19 hours ago



















0












$begingroup$

Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
$$y^2=9-frac{9}{25}x^2$$ you will have the objective function
$$f(x)=frac{16}{25}x^2+9$$






share|cite|improve this answer











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$



    Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
    $$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$



    The second gives a minimal value, wile the first gives a maximal value:
    $$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$



    If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$



    and we'll get the same answer of course.



    It happens because $-frac{1}{9}cdot225=-25$ and
    $$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
      $endgroup$
      – PGupta
      21 hours ago










    • $begingroup$
      @PGupta I added something. See now.
      $endgroup$
      – Michael Rozenberg
      21 hours ago










    • $begingroup$
      This makes sense, thank you.
      $endgroup$
      – PGupta
      21 hours ago
















    2












    $begingroup$

    Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$



    Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
    $$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$



    The second gives a minimal value, wile the first gives a maximal value:
    $$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$



    If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$



    and we'll get the same answer of course.



    It happens because $-frac{1}{9}cdot225=-25$ and
    $$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
      $endgroup$
      – PGupta
      21 hours ago










    • $begingroup$
      @PGupta I added something. See now.
      $endgroup$
      – Michael Rozenberg
      21 hours ago










    • $begingroup$
      This makes sense, thank you.
      $endgroup$
      – PGupta
      21 hours ago














    2












    2








    2





    $begingroup$

    Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$



    Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
    $$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$



    The second gives a minimal value, wile the first gives a maximal value:
    $$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$



    If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$



    and we'll get the same answer of course.



    It happens because $-frac{1}{9}cdot225=-25$ and
    $$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$






    share|cite|improve this answer











    $endgroup$



    Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$



    Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
    $$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$



    The second gives a minimal value, wile the first gives a maximal value:
    $$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$



    If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$



    and we'll get the same answer of course.



    It happens because $-frac{1}{9}cdot225=-25$ and
    $$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 21 hours ago

























    answered 21 hours ago









    Michael RozenbergMichael Rozenberg

    110k1896201




    110k1896201












    • $begingroup$
      Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
      $endgroup$
      – PGupta
      21 hours ago










    • $begingroup$
      @PGupta I added something. See now.
      $endgroup$
      – Michael Rozenberg
      21 hours ago










    • $begingroup$
      This makes sense, thank you.
      $endgroup$
      – PGupta
      21 hours ago


















    • $begingroup$
      Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
      $endgroup$
      – PGupta
      21 hours ago










    • $begingroup$
      @PGupta I added something. See now.
      $endgroup$
      – Michael Rozenberg
      21 hours ago










    • $begingroup$
      This makes sense, thank you.
      $endgroup$
      – PGupta
      21 hours ago
















    $begingroup$
    Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
    $endgroup$
    – PGupta
    21 hours ago




    $begingroup$
    Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
    $endgroup$
    – PGupta
    21 hours ago












    $begingroup$
    @PGupta I added something. See now.
    $endgroup$
    – Michael Rozenberg
    21 hours ago




    $begingroup$
    @PGupta I added something. See now.
    $endgroup$
    – Michael Rozenberg
    21 hours ago












    $begingroup$
    This makes sense, thank you.
    $endgroup$
    – PGupta
    21 hours ago




    $begingroup$
    This makes sense, thank you.
    $endgroup$
    – PGupta
    21 hours ago











    4












    $begingroup$

    If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
    $$
    tag{1} nabla f(x^star) = lambda nabla g(x^star).
    $$
    If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.



    Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
      $endgroup$
      – PGupta
      21 hours ago






    • 1




      $begingroup$
      @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
      $endgroup$
      – littleO
      20 hours ago










    • $begingroup$
      Great, got it. Thanks a lot!
      $endgroup$
      – PGupta
      19 hours ago
















    4












    $begingroup$

    If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
    $$
    tag{1} nabla f(x^star) = lambda nabla g(x^star).
    $$
    If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.



    Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
      $endgroup$
      – PGupta
      21 hours ago






    • 1




      $begingroup$
      @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
      $endgroup$
      – littleO
      20 hours ago










    • $begingroup$
      Great, got it. Thanks a lot!
      $endgroup$
      – PGupta
      19 hours ago














    4












    4








    4





    $begingroup$

    If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
    $$
    tag{1} nabla f(x^star) = lambda nabla g(x^star).
    $$
    If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.



    Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.






    share|cite|improve this answer









    $endgroup$



    If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
    $$
    tag{1} nabla f(x^star) = lambda nabla g(x^star).
    $$
    If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.



    Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 21 hours ago









    littleOlittleO

    30.4k648111




    30.4k648111












    • $begingroup$
      Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
      $endgroup$
      – PGupta
      21 hours ago






    • 1




      $begingroup$
      @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
      $endgroup$
      – littleO
      20 hours ago










    • $begingroup$
      Great, got it. Thanks a lot!
      $endgroup$
      – PGupta
      19 hours ago


















    • $begingroup$
      Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
      $endgroup$
      – PGupta
      21 hours ago






    • 1




      $begingroup$
      @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
      $endgroup$
      – littleO
      20 hours ago










    • $begingroup$
      Great, got it. Thanks a lot!
      $endgroup$
      – PGupta
      19 hours ago
















    $begingroup$
    Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
    $endgroup$
    – PGupta
    21 hours ago




    $begingroup$
    Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
    $endgroup$
    – PGupta
    21 hours ago




    1




    1




    $begingroup$
    @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
    $endgroup$
    – littleO
    20 hours ago




    $begingroup$
    @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
    $endgroup$
    – littleO
    20 hours ago












    $begingroup$
    Great, got it. Thanks a lot!
    $endgroup$
    – PGupta
    19 hours ago




    $begingroup$
    Great, got it. Thanks a lot!
    $endgroup$
    – PGupta
    19 hours ago











    0












    $begingroup$

    Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
    $$y^2=9-frac{9}{25}x^2$$ you will have the objective function
    $$f(x)=frac{16}{25}x^2+9$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
      $$y^2=9-frac{9}{25}x^2$$ you will have the objective function
      $$f(x)=frac{16}{25}x^2+9$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
        $$y^2=9-frac{9}{25}x^2$$ you will have the objective function
        $$f(x)=frac{16}{25}x^2+9$$






        share|cite|improve this answer











        $endgroup$



        Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
        $$y^2=9-frac{9}{25}x^2$$ you will have the objective function
        $$f(x)=frac{16}{25}x^2+9$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 21 hours ago

























        answered 22 hours ago









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        78.7k42867




        78.7k42867






























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